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16 tháng 2 2019

B,x(x-1)(x-2)(x-3)=15

(x2-3x)(x2-3x+2)=15

Đặt x2-3x+1=k

(k-1)(k+1)=15

k2=16

\(\orbr{\begin{cases}k=4\\k=-4\end{cases}}\)

\(\orbr{\begin{cases}x^2-3x+1=4\\x^2-3x+1=-4\end{cases}}\)

Vậy pt vô nghiệm

k nhá

16 tháng 2 2019

ê bạn j đó ơi 

chỗ thứ 2 bn ghép kiểu gì vậy 

12 tháng 2 2017

\(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)

\(\Leftrightarrow\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x-3-2x+5\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(2-x\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}3x+1=0\\2-x=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[\begin{matrix}3x=-1\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=-\frac{1}{3}\\x=2\end{matrix}\right.\)

Vậy tập nghiệm của pt là \(S=\left\{-\frac{1}{3};2\right\}\)

12 tháng 2 2017

Có : \(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)

\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3-2x+5\right)=0\)

\(\Leftrightarrow\) \(\left(3x+1\right)\left(-x+2\right)=0\)

\(\Leftrightarrow\) \(\left[\begin{matrix}3x+1=0\\-x+2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}3x=-1\\-x=-2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}x=\frac{-1}{3}\\x=2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{-1}{3};2\right\}\)

1 tháng 4 2019

a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(-x^2+2x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)

1 tháng 4 2019

(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)

\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)

\(-x^3-x^2+9x+9=0\)

\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)

\(\left(x+1\right)\left(9-x^2\right)\)=0

(x+1)(3-x)(3+x)=0

*x+1=0 =>x=-1

*3-x=0=>x=3

*3+x=0=>x=-3

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

8 tháng 2 2020

Câu 1 :

8 tháng 2 2020

a, Ta có : \(3\left(x-1\right)-2\left(x+3\right)=-15\)

=> \(3x-3-2x-6=-15\)

=> \(3x-3-2x-6+15=0\)

=> \(x=-6\)

Vậy phương trình có nghiệm là x = -6 .

b, Ta có : \(3\left(x-1\right)+2=3x-1\)

=> \(3x-3+2=3x-1\)

=> \(3x-3+2-3x+1=0\)

=> \(0=0\)

Vậy phương trình có vô số nghiệm .

c, Ta có : \(7\left(2-5x\right)-5=4\left(4-6x\right)\)

=> \(14-35x-5=16-24x\)

=> \(14-35x-5-16+24x=0\)

=> \(-35x+24x=7\)

=> \(x=\frac{-7}{11}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{11}\) .

Bài 2 :

a, Ta có : \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)

=> \(\frac{x}{30}+\frac{3\left(5x-1\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{5\left(2x+3\right)}{30}\)

=> \(x+3\left(5x-1\right)=2\left(x-8\right)-5\left(2x+3\right)\)

=> \(x+15x-3=2x-16-10x-15\)

=> \(x+15x-3-2x+16+10x+15=0\)

=> \(24x+28=0\)

=> \(x=\frac{-28}{24}=\frac{-7}{6}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{6}\) .

b, Ta có : \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

=> \(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)

=> \(6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)

=> \(6x+24-30x+120=10x-15x+30\)

=> \(6x+24-30x+120-10x+15x-30=0\)

=> \(-19x+114=0\)

=> \(x=\frac{-114}{-19}=6\)

Vậy phương trình có nghiệm là x = 6 .

a) \(15x-3\left(3x-2\right)=45-5\left(2x-5\right)\)

\(\Leftrightarrow15x-9x+6=45-10x+25\)

\(\Leftrightarrow15x-9x+10x=45+25-6\)

\(\Leftrightarrow16x=64\)

\(\Leftrightarrow x=4\)

b) \(x^2-9+4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)+4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\x+7=0\Leftrightarrow x=-7\end{matrix}\right.\)

c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)

\(\Leftrightarrow\dfrac{x+4+\left(x+2\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\)

\(\Leftrightarrow x+4+x^2-4x+2x-8=5x-4\)

\(\Leftrightarrow x^2+x-4x+2x-5x=-4+8-4\)

\(\Leftrightarrow x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\Leftrightarrow x=6\end{matrix}\right.\)

12 tháng 4 2018

a) 15x - 3(3x - 2) = 45 - 5(2x - 5)

\(\Leftrightarrow\) 15x - 9x + 6 = 45 - 10x + 25

\(\Leftrightarrow\) 6x + 10x = 70 - 6

\(\Leftrightarrow\) 16x = 64

\(\Leftrightarrow\) x = 4

Vậy.......................

b) x2 - 9 + 4(x - 3) = 0

\(\Leftrightarrow\) (x - 3)(x + 3) + 4(x - 3) = 0

\(\Leftrightarrow\) (x - 3)(x + 3 + 4) = 0

\(\Leftrightarrow\) (x - 3)(x + 7) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=3\end{matrix}\right.\)

Vậy........................

c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)

\(\Leftrightarrow\) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\) (đk: x\(\ne\pm\)4)

\(\Leftrightarrow\) \(\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}+\dfrac{\left(x+2\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5x-4}{\left(x+4\right)\left(x-4\right)}\)

\(\Leftrightarrow\) x + 4 + x2 - 4x + 2x - 8 = 5x - 4

\(\Leftrightarrow\) x2 - x - 5x - 4 + 4 = 0

\(\Leftrightarrow\) x2 - 6x = 0

\(\Leftrightarrow\) x(x - 6) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=6\left(tmđk\right)\end{matrix}\right.\)

Vậy...............

23 tháng 4 2020

Bài 1:

a, \(\frac{1}{x+1}+\frac{2}{x-1}=\frac{1+x^2}{x^2-1}\) (ĐKXĐ: x \(\ne\) \(\pm\) 1)

\(\Leftrightarrow\) \(\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{1+x^2}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow\) x - 1 + 2(x + 1) = 1 + x2

\(\Leftrightarrow\) x - 1 + 2x + 2 - 1 - x2 = 0

\(\Leftrightarrow\) -x2 + 3x = 0

\(\Leftrightarrow\) x(3 - x) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐKXĐ\right)\\x=3\left(TMĐKXĐ\right)\end{matrix}\right.\)

Vậy S = {0; 3}

b, \(\frac{x-2}{x+2}-\frac{x}{x-2}=\frac{8}{x^2-4}\) (ĐKXĐ: x \(\ne\) \(\pm\) 2)

\(\Leftrightarrow\) \(\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{8}{\left(x+2\right)\left(x-2\right)}\)

\(\Rightarrow\) (x - 2)2 - x(x + 2) = 8

\(\Leftrightarrow\) (x - 2)2 - x(x + 2) - 8 = 0

\(\Leftrightarrow\) x2 - 4x + 4 - x2 - 2x - 8 = 0

\(\Leftrightarrow\) -6x - 4 = 0

\(\Leftrightarrow\) x = \(\frac{-2}{3}\) (TMĐKXĐ)

Vậy S = {\(\frac{-2}{3}\)}

c, \(\frac{1}{x}\) + \(\frac{2}{x-3}\) = \(\frac{1-5x}{x^2-3x}\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) 3)

\(\Leftrightarrow\) \(\frac{x-3}{x\left(x-3\right)}+\frac{2x}{x\left(x-3\right)}=\frac{1-5x}{x\left(x-3\right)}\)

\(\Rightarrow\) x - 3 + 2x = 1 - 5x

\(\Leftrightarrow\) 3x - 3 = 1 - 5x

\(\Leftrightarrow\) 3x + 5x = 1 + 3

\(\Leftrightarrow\) 8x = 4

\(\Leftrightarrow\) x = \(\frac{1}{2}\) (TMĐKXĐ)

Vậy S = {\(\frac{1}{2}\)}

Bài 2:

a, \(\frac{1}{x+2}=\frac{5}{2-x}+\frac{12+x}{x^2-4}\)

\(\Leftrightarrow\) \(\frac{1}{x+2}=\frac{-5}{x-2}+\frac{12+x}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\) \(\frac{x-2}{\left(x+2\right)\left(x-2\right)}=\frac{-5\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12+x}{\left(x+2\right)\left(x-2\right)}\)

\(\Rightarrow\) x - 2 = -5(x + 2) + 12 + x

\(\Leftrightarrow\) x - 2 = -5x - 10 + 12 + x

\(\Leftrightarrow\) x - 2 = -4x + 2

\(\Leftrightarrow\) x + 4x = 2 + 2

\(\Leftrightarrow\) 5x = 4

\(\Leftrightarrow\) x = \(\frac{4}{5}\)

Vậy S = {\(\frac{4}{5}\)}

Chúc bn học tốt!! (Phần b hình như không có gì thì phải)

22 tháng 4 2022

Bài 1: 

c) |2x - 1| = x + 2

<=> 2x - 1 = +(x + 2) hoặc -(x + 2)

* 2x - 1 = x + 2      

<=> 2x - x = 2 + 1

<=> x = 3

* 2x - 1 = -(x + 2)

<=> 2x - 1 = x - 2

<=> 2x - x = -2 + 1

<=> x = -1

Vậy.....

1: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)

\(\Leftrightarrow\dfrac{5x^2-12}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x^2-5x}{\left(x+1\right)\left(x-1\right)}\)

Suy ra: \(5x^2+3x-9=5x^2-5x\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(tm\right)\)

2: Ta có: \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)

\(\Leftrightarrow\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3x-15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)

Suy ra: \(6x=3x-15\)

\(\Leftrightarrow3x=-15\)

hay \(x=-5\left(loại\right)\)

 

AH
Akai Haruma
Giáo viên
19 tháng 8 2021

2. ĐKXĐ: $x\neq \pm 5$
PT \(\Leftrightarrow \frac{3}{x-5}+\frac{3x-15}{x^2-25}=\frac{3}{x+5}\)

\(\Leftrightarrow \frac{3}{x-5}+\frac{3(x-5)}{(x-5)(x+5)}=\frac{3}{x+5}\)

\(\Leftrightarrow \frac{3}{x-5}+\frac{3}{x+5}=\frac{3}{x+5}\Leftrightarrow \frac{3}{x-5}=0\) (vô lý)

Vậy pt vô nghiệm.